# Relating linear and angular kinematics

kenren
im really having a hard tym doing this assignment of mine, i hope someone cud help me out on this one.

at t= 3.00s a point on the rim of a 0.200-m-radius wheel has a tangential speed of 50.0 m/s as the whell slows down with a tangential acceleration of constant magnitude 10.0 m/s2.
a.)Calculate the wheel's constant angular acceleration.
b.) calculate the wheel's constant angular velocities at t= 3.00s and t= 0.
c.) through what angle did the wheel turn between t=0 and t=3.00s?
At what time will the radial acceleration equal g?

can you tell me the formula and the proccess on how to solve this problems..please.. ive tried to solve this one but just couldnt get any good answers. please neone help me.... Last edited by a moderator:

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HallsofIvy
Homework Helper
An entire circle is 2pi radians and the circumference of a circle of radius 0.2 m is .4 pi m. That means that an angular velocity of omega radians/s is the same as (1/(2pi)omega) revolutions per second or (.4pi)(1/(2pi))omega = .2 omega m/s tangential speed. Since you are given that, at t=3, the tangential speed is 50 m/s, it follow that the angular velocity at that time satisfies 50= .2 omega or
omega= 50/.2= 250 radians/s. (That's part of the answer to part b.)

If the wheel was slowing down at a constant 10 m/s^2, then, when t=0, the tangential speed must have been 50+ 3(10)= 80 m/s^2. By the same argument, the angular velocity at that time was 80/.2=

Since the angular velocity changed from 400 radians/s to 250 radians/s in 3 seconds, the angular acceleration is (250-400)/3=